If $f(x)$ is integrable can we say that $f(x)^n$ is integrable?

Question

Suppose $f(x)$ is a positive and continuously differentiable functions. In addition, it is well-known that $\int_{0}^{\infty} f(x)dx$ is bounded. My point of view is that $\int_{0}^{\infty} f(x)^mdx$ (where $m \in \mathbb N$) is bounded. I will be gratefull if you would propose me a conterexample. If this claim is true, how can I prove it?

Answer 1

Define the following function: $$f(x)=\begin{cases} 2/\sqrt{x}-1,&0<x\leq1 \\ e^{1-x},&x>1. \end{cases}$$ Then $f$ is continuously differentiable, and we have: $$\int_0^\infty f(x)\,dx=\int_0^1\frac{2}{\sqrt{x}}-1\,dx+\int_1^\infty e^{1-x}\,dx=4,$$ but: $$\int_0^\infty f^2(x)\,dx=\int_0^1 1-\frac{4}{\sqrt x}+\frac{4}{x}\,dx+\int_1^\infty e^{2-2x}\,dx,$$ and $\displaystyle\int_0^11-\frac{4}{\sqrt x}+\frac{4}{x}\,dx$ diverges.

Answer 2

Even if $f(x)$ is continuous at $x=0$, it is not necessarily true that the convergence of $\int_0^\infty f(x)\,dx$ implies the convergence of $\int_0^\infty f(x)^m\,dx$. The following is a counterexample for every $m\ge2$:

Let $I_1,I_2,I_3,\dots$ be a sequence of disjoint intervals of lengths $\frac1{1^4},\frac1{2^4},\frac1{3^4},\dots$. Define $$ f(x) = \begin{cases} j^2, &\text{if $x\in I_j$ for some $j\ge1$}, \\ 0, &\text{otherwise}. \end{cases} $$ Then $\int_0^\infty f(x)\,dx = \sum_{j=1}^\infty j^2 \frac1{j^4} = \frac{\pi^2}6 < \infty$, but $\int_0^\infty f(x)^2\,dx = \sum_{j=1}^\infty j^4 \frac1{j^4} = \infty$.

Admittedly this $f(x)$ is not continuous, but it is easy to approximate it arbitrarily closely (in the $L^1$ and $L^2$ means) by arbitrarily smooth functions, which will still serve as counterexamples.

If one adds the additional assumption that the positive function $f(x)$ tends to $0$ as $x\to\infty$, then the convergence of $\int_0^\infty f(x)\,dx$ does imply the convergence of $\int_0^\infty f(x)^m\,dx$ (even without an assumption of continuity or differentiability).